The standard electrode potential ( $\mathrm{E}^{\circ}$ ) for the half-cell reaction $\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}$ at 298 K is (Given : $\mathrm{E}^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}\right)=-0.04 \mathrm{~V}$ and $\mathrm{E}^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.44 \mathrm{~V}$ at 298 K )

For a salt XY, which is a strong electrolyte, the plot of $\Lambda_{\mathrm{m}}$ versus $\sqrt{\mathrm{c}}$ has a slope of $-90.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{L}^{1 / 2}$ at 298 K . At 0.01 M concentration of $\mathbf{X Y}$, the value of $\Lambda_{\mathrm{m}}$ is $145.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The limiting molar conductivity of $\mathbf{Y}^{-}$ion $\left(\lambda_{\mathbf{Y}^{-}}^0\right.$, in $\left.\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1}\right)$ at 298 K will be

(Given : $\lambda_{\mathrm{X}^{+}}^0=74.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ )

A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is :

(Given : Molar mass of $\mathrm{Cu}=63 \mathrm{~g} \mathrm{~mol}^{-1}$;

$$ \left.1 \mathrm{~F}=96487 \mathrm{C} \mathrm{~mol}^{-1}\right) $$

Calculate emf of the half cell given below :

$$ \begin{aligned} & \mathrm{Pt}(\mathrm{~s})\left|\mathrm{H}_2(\mathrm{~g}, 2 \mathrm{~atm})\right| \mathrm{HCl}(\mathrm{aq}, 0.02 \mathrm{M}) \\ & \mathrm{E}_{\mathrm{H}_2 / \mathrm{H}^{+}}^{\circ}=0 \mathrm{~V} \end{aligned} $$

(Given: $\frac{2.303 R T}{F}=0.059, \log 2=0.3010$ )